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I: X'->X the identity function with topology

  1. Feb 23, 2009 #1
    Let X and X' denote a single set in the two topologies T and T', respectively. Let i:X'-> X be the identity function
    a. Show that i is continuous <=> T' is finer than T.
    b. Show that i is a homeomorphism <=> T'=T

    This is all i've got.

    According to the first statement... X [tex]\subset[/tex] T and X' [tex]\subset[/tex] T

    a. if i is continuous... each open subset V of X' the set i^-1 is an open subset of X
    T' is finer than T means... T [tex]\subset[/tex] T'.

    i don't know where to go from here...

    b. if i is a homeomorphism...
    then... i is... a bijection therefore the function and the inverse function are continuous.
    i and i^-1 are continuous. each open subset V of X' the set i^-1 is an open subset of X
    if T' = T... then.....

    i don't know where to go from here either...

    can someone help me out?

    Thank You,
    tomboi03
     
  2. jcsd
  3. Feb 24, 2009 #2

    lanedance

    User Avatar
    Homework Helper

    don't know a heap about this stuff but would think you were pretty close, will have a try...

    so
    X [tex]\in[/tex] T
    X' [tex]\in[/tex] T'

    for a)
    prove ->

    i is the identity X' -> X
    continuous means preimage of an open set is open

    so given V[tex]\in[/tex] T (as the topology contains all the open sets)
    then V' = i-1.V is open
    so V' is in T'
    but
    V' = i.V' = i. i-1.V = V (little sketchy about this part??)
    so
    if
    V[tex]\in[/tex] T
    then
    V is in T'

    so T [tex]\subseteq[/tex] T'

    now need to prove other direction....
     
    Last edited: Feb 24, 2009
  4. Feb 24, 2009 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    T' "finer" than T means that any open set in T is an open set in T' (but there may be open sets in T' that are not in T).

    Suppose T' is finer than T. If U is an open set in T, then i-1 maps it back to itself, which is an open set in T' so the identity map is continuous.

    Suppose i is continuous. It U is an open set in T, then i-1 maps it back to itself, which, because i is continuous, is open and so in T'. Thus, every member of T is in T' and so T' is finer than T.
     
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